# How to show the following equality of ordinal exponentiation: $n^{\omega^\omega}=\omega^{\omega^\omega}$?

Let $n > 1$ be a positive integer. Prove that $n^{\omega^\omega}=\omega^{\omega^\omega}$

I was able to prove that, since $ω$ is a limit ordinal, and $n > 1$, $$n^\omega= \bigcup_{\beta<\omega}n^\beta = \omega$$ Therefore I end up with $n^{\omega^\omega}=\omega^\omega$, and am left to prove that $\omega^\omega=\omega^{\omega^\omega}$ but that is something that I do not believe holds.

-
I've noticed that you have asked quite a few questions recently. I wanted to make sure that you are aware of the quotas 50 questions/30 days and 6 questions/24 hours, so that you can plan posting your questions accordingly. (If you try to post more questions, stackexchange software will not allow you to do so.) For more details see meta. – Asaf Karagila Nov 25 '12 at 18:42
Is this an ordinal or cardinal exponentiation? It makes a huge difference. Also, note that exponentiation is not associative. Please use parenthesis to clarify. – Asaf Karagila Nov 25 '12 at 18:43
I changed the writing the best I could to clarify what I am trying to say. I am trying to show that n to the omega to the omega is equal to omega to the omega to the omega. – Omar Nov 25 '12 at 19:07
I am not given any specification of whether this is ordinal or cardinal exponentiation, but I do believe that it is ordinal. – Omar Nov 25 '12 at 19:07

Note that $(n^\omega)^\omega$ is not $n^{\omega^\omega}$. Your calculation was of the former, which is indeed different from $\omega^{\omega^\omega}$.

Let us calculate what $n^{\omega^\omega}$ is. As you know, $\omega^\omega$ is a limit ordinal, so we have that $$n^{\omega^\omega}=\sup\{n^\gamma\mid \gamma<\omega^\omega\}$$

It is enough to calculate this supremum over a cofinal sequence, e.g. $\{\omega^k\mid k\in\omega\}$. So we are suppose to calculate the supremum of $n^{\omega^k}$ for $k\in\omega$. But $\omega^k$ is also a limit ordinal, we once again we unwind the definitions: $$n^{\omega^k} = \sup\{n^\gamma\mid\gamma<\omega^k\}$$

Continuing this unwinding we see that $\omega^k$ is the supremum of $\omega^{k-1}\cdot m$ where $m\in\omega$. But that too is a limit ordinal, so we once again have to find a cofinal sequence $\omega^{k-1}\cdot(m-1)+t$.

So we have that $n^{\omega^1}=\omega$ as you said, and by exponentiation laws we have $n^{\omega^k}=\omega^{\omega^{k-1}}$, but this is really what because now: $$\sup\{n^{\omega^k}\mid k\in\omega\}=\sup\{\omega^{\omega^{k-1}}\mid k\in\omega, k>0\}=\sup\{\omega^{\omega^k}\mid k\in\omega\}=\omega^{\omega^\omega}$$

-
I know this seem an irrelevant question, I just wanted to make sure that "sup" was the same as "max" . My teacher has never used "sup" and I wanted to make sure there wasn't any kind of difference between the two. I am also confused on how an infinite set can have a least upper bound? – Omar Nov 26 '12 at 19:47
@Kyle: The supremum of a set of ordinals is its union. It is the least ordinal which is greater or equal than the elements of the set. The supremum itself is not taken within the set but rather in the class of ordinals. E.g. $$\omega=\sup\{\alpha\mid\alpha\text{ is a finite ordinal}\}$$ While the set of finite ordinals itself is unbounded $\omega$ is its supremum within the class of ordinals. – Asaf Karagila Nov 26 '12 at 19:59